Integration 2

$\begin{aligned} I & = \int \csc^2 x \ dx \\ & = \int \frac 1{\sin^2 x} \ dx & \small \color{#3D99F6} \text{By half-angle tangent substitution and} \\ & = \int \frac 1{\left(\frac {2t}{1+t^2}\right)^2} \cdot \frac 2{1+t^2} dt & \small \color{#3D99F6} \text{let } t = \tan \frac x2 \implies dx = \frac {2\ dt}{1+t^2} \\ & = \int \frac {1+t^2}{2t^2} \ dt \\ & = \frac 12 \int \left(\frac 1{t^2} + 1\right) dt \\ & = \frac 12 \left(-\frac 1t + t\right) + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = - \frac {1-t^2}{2t} + C \\ & = \boxed{- \cot x + C} \end{aligned}$

Of course, if you already know the derivative of $\cot(x)$ (which is a problem that a student will likely explore in a first course in calculus), then that pretty much gives you the answer to this question. And even if you didn't already know that, you could've just looked at the answer choices, and started taking derivatives. Going based off the assumption, though, that you do not recognize the integrand as the derivative of $(-\cot{x})$ and that you do not have any answer choices, here is an alternative method other than Chew-Seong Cheong's. (And I should add that Chew-Seong Cheong uses a powerful substitution choice that is worth remembering that I first learned on brilliant.org, sometimes called the Weierstrass substitution.) I present two different solutions since the first solution uses the fact that $\int \sec^2{x}dx=\tan{x}+C$ , which is a similar integral to the original. $$ Solution 1: $\int \csc^2{x}dx= \int \frac{\sec^2{x}}{\tan^2{x}}dx=\int \frac{d(\tan{x})}{\tan^2{x}}=-\frac{1}{\tan{x}}+C=-\cot{x}+C$ Solution 2: $\int \csc^2{x}dx= \int \csc^2{x}\frac{dx}{d(\csc^2{x})}d(\csc^2{x}) =\int \csc^2{x}\Big(\frac{d(\csc^2{x})}{dx}\Big)^{-1}d(\csc^2{x}-1)=\int \csc^2{x}(-2\csc^2{x}\cot{x})^{-1}d(\cot^2{x})$ $=\int \frac{-1}{2\cot{x}}d(\cot^2{x})=\int \frac{-1}{2\sqrt{\cot^2{x}}}d(\cot^2{x})=-\sqrt{\cot^2{x}}+C=-\cot{x}+C$